A long spring when stretched by `x cm`, has a potential energy `U`. On increasing the stretching to `nx cm,` the potential energy stored in spring will be

Options

A. `(U)/(n)`
B. `nU`
C. `n^(2)U`
D. `(U)/(n^(2))`

Answer

C

Explanation

Potential energy of spring
`U=(1)/(2)kx^(2)`
`U'=(1)/(2)K(nx)^(2)impliesn^(2)[(1)/(2)Kx^(2)]`
`U'=n^(2)U`